Denote \(\mathring{C}\)  as the interior of \(C\) . The fact that \(\ell (\mathcal {Z}\cap \mathring{C})\)  has an empty interior is a consequence of Sard's theorem and the fact that \(\ell \)  is \(d\) -times differentiable (see [1]}). We now show that the image of \(\ell \)  of any \(m\) -dimensional boundary of \(C\)  intersected by \(\mathcal {Z}\)  also has an empty interior. Consider \(m > 0\) , and fix \(m-d\)  coordinates of \(C\)  as \(c_{m+1}, \dots , c_{d}\) , where \(c_i\)  is equal to \(a_i\)  or \(b_i\) , and denote \(C_m = (a_1, b_1) \times (a_2, b_2) \dots \times (a_m, b_m) \times \lbrace  c_{m+1}\rbrace  \times \dots \times \lbrace c_{d}\rbrace \) . Note that if \(y \in \mathcal {Z}\cap C_m\) , then the \(m\)  first coordinates of \(\nabla \ell (y)\)  are zero. Thus, if we call \(\ell _m\)  the restriction of \(\ell \)  to \(C_m\) , then \(\ell _m : C_m \rightarrow \mathbb {R}\)  is \(d\)  times differentiable and \(\mathcal {Z}\cap C_m\)  is included in its set of critical points. Applying Sard's theorem to \(\ell _m\) , we obtain that \(\ell (\mathcal {Z}\cap C_m)\)  has an empty interior. Since \(C\)  can be written as a union of these \(C_m\) , this completes the proof.
